Question

An electric field of magnitude $1000\text{N/C}$ is produced between two parallel plates having a separation of $2.0\text{cm}$ as shown in figure. With what minimum speed should an electron be projected from the lower plate in the direction of the field so that it may just reach the upper plate ?
(Assume gravity free space)
(Charge on electron is $-1.6\times {10}^{-19}\text{C}$ and mass of electron is $9.1\times {10}^{-31}\text{kg}$)

• A. $1.7\times {10}^6\text{m/s}$
• B. $2.7\times {10}^6\text{m/s}$
• C. $3.7\times {10}^6\text{m/s}$
• D. $4.7\times {10}^6\text{m/s}$

Answer 1

The correct option is B $2.7\times {10}^6\text{m/s}$

Given that,
Magnitude of electric field, $E=1000\text{N/C}$
Charge of an electron, $q=-1.6\times {10}^{-19}\text{C}$
Mass of an electron, $m=9.1\times {10}^{-31}\text{kg}$


Force on the electron, $F=qE$ (in downward direction)
$\Rightarrow F=1.6\times {10}^{-19}\times 1000$
$\Rightarrow F=1.6\times {10}^{-16}\text{N}$ (retarding)
Acceleration, $a=\frac{F}{m}$
$\Rightarrow a=\frac{1.6\times {10}^{-16}}{9.1\times {10}^{-31}}=1.8\times {10}^{14}{\text{m/s}}^2$ (retarding)
Let us suppose the electron is projected with speed $u$.
So, $v^2-u^2=2as$
$\Rightarrow 0-u^2=-2\times 1.8\times {10}^{14}\times 2\times {10}^{-2}$ [to just reach, $v=0$]
$\Rightarrow u=\sqrt{2\times 1.8\times {10}^{14}\times 2\times {10}^{-2}}$
$\Rightarrow u=2.7\times {10}^6\text{m/s}$